<snip>>>>>> Here, the conundrum is that in ZFC there are uncountably many>>> irrationals, that there be uncountably many disjoint intervals, that>>> each contains a rational, which are countable in ZFC, which would be a>>> contradiction.>>>> ZFC does not say anything about the topology of the>> real line.>>>> There are better topologists out here than I. But, while>> the reals are not compact, they are sigma compact. Or, if>> I have that wrong, they have whatever compactness condition>> is required so that the second-countability of the space>> applies to your example.>>>> The identity criterion in set theory is not the identity>> criterion of the real numbers. The latter come from the>> topological completeness by which is meant completeness>> relative to the rationals. Thus, trichotomy of the>> rationals yields trichotomy in the reals. Hence, the>> reals have an identity criterion based on their order>> relation>>>> x=y <-> (x<=y /\ x>=y)>>>> If you want to think about how the identity criterion>> of a logical system like ZFC is related to topology>> look at the difference between the axioms of a metric>> and a pseudometric. For one you have,>>>> x=y<->d(x,y)=0>>>> and for the other you have>>>> x=y->d(x,y)=0>>>> The latter statement is what is involved with attaching>> metric structure (and, therefore, a metric based>> topology) to a logical system.>>>> The next step is to look at uniform spaces defined>> relative to uniformities. That will give the first>> conditions for a system of relations to have>> a topology that may be metrizable.>>>> Finally, find a proof for the metrization of relations>> so that you can see what is involved with attaching a>> pseudometric to a logical system. You can find one>> in Kelley's "General Topology".>>>> Curiously, it is circular when speaking to the>> metrization of the reals. It invokes the least>> upper bound property. Still, it is instructive.>> These topological properties of density in the reals of the> irrationals and rationals build up from the density of the rationals,> the non-continuity of the rationals, and the irrationals being the> complement of the rationals, in the reals.>

Starting here, I have not yet read what follows.

However, there is an issue of logical priority ofdefinition. The reals are defined as a logical type.In the case of Dedekind, it was cuts. In the case ofCantor it was fundamental sequences.

That is why the identity criterion of the real numbersfollows from the trichotomy on the rationals. It is nota topological property in the sense of the usual topologyfollowing from open sets defined by a metric.

In other words, the irrationals are not simply thecomplement of the rationals. They are infinite setsof rationals that have no limit point in the rationals.

They are only the complement of "rationals" that havebeen defined as infinite sets of rationals having alimit point that is a rational.

> The cardinal properties of the rationals and irrationals (here often> specifically for the unit interval instead of the entire set) follow> from ZFC and cardinality of the reals, and the general definition of> the rationals as integer ratios and reals as sufficiently represented> as infinite expansions or as Cauchy or as equivalence classes of> Cauchy sequences.

I disagree that the general definition of therationals in this context are integer ratios.

In the definition of real numbers, the usual sequenceof logical types takes the natural numbers as given:

0, 1, 2, 3, ...

Integers are formed as infinite sets of ordered pairssuch that "equivalent pairs have the same difference"

(2,3) EQ (12,13)

-((2,3) EQ (3,2))

Rationals are formed according to the same trick.You are correct, ratios are involved. But each rationalis an infinite collection of ordered pairs of integers.

There are no rational numbers in ZFC. There is, however,a representation of the rationals along the lines givenhere.

In each step, the order of the new logical type isrelated to the prior logical type.

So, since 0<2 in the natural numbers, observethat the order in the ordered pair

(0,2)

is coherent with the order of the statement,let it be a representative from the equivalenceclass that shall correspond with +2. By constrast,let

(2,0)

be a representative from the equivalence classthat shall correspond with -2.

For a proper fraction greater than zero, thenumerator is strictly smaller than the denominator.So, let

((0,2),(0,3))

be a representative of the equivalence classthat shall correspond with 2/3 and let

((0,3),(0,2))

be a representative of the equivalence classthat shall correspond with 3/2. Then eitherof

((2,0),(0,3)) or ((0,2),(3,0))

are representatives of the equivalenceclass for -2/3.

It has been a long time since I actuallylooked at published version of thisconstruction. I may have gotten partof it wrong (negative rationals?). But,this should give you the general ideaof my objection to what you haveposited as the general definitionof the rationals.

This does not mean I am correct. Justthat we see things differently as tohow the rationals are understood inZFC.

>> I'm not familiar yet with "second-countability". It might seem from> its statement an order type property, of the rationals as having order> type 2^w. (As would (0,1)^w have were it countable.)>> http://en.wikipedia.org/wiki/Second-countable_space>> Second-countability is a property of a space that it is completely> separable, that the topology of the space has a basis that is> countable. This basically has that there aren't any open subsets that> aren't countable unions of some countable collection of open> subsets.

I do not believe I have ever seen anyone talkabout a topology in terms of order type unlessthey were talking specifically about a topologyon the ordinals themselves.

A topology is simply a partition of the power setfor a given set that includes the null class andthe given set itself among the members. Onecan either call the sets open or closed anddepending on that designation, they must satisfycertain axioms. For open sets,

1. Every arbitrary union of open sets is open.

2. Every finite intersection of open sets is open.

A basis for open sets is a collection of opensets such that for any given open set and anygiven element of that open set, thereis a basis open set that is a subset of thegiven open set that also has the givenelement as an element.

Because the rationals are countable anddense, open balls defined by the metricby rational numbers

{y|d(x,y)<m/n, m,n integers}

are a countable basis for the topology.

>> Then, while there's a countable cover of open subsets of R, here of> intervals, the point is that there are uncountably many points in R,> in ZFC, each defines an interval, closed with endpoints open without,> of the set between zero and it. Where these are the irrationals then> R though completely separable, _does_ have uncountable unions of open> sets the union of which may be an open set that is not the entire> set. It still may only take a countable union of open subsets to form> any open subset, but, open subsets of irrationals, can also be> considered uncountable unions of open sets, bounded by an irrational> but not rationals, simply as they would be. Similarly closed sets> bounded on at least one side by an irrational may be the union of> uncountably many closed sets, bounded by an irrational but not> rationals (as there aren't uncountably many rationals).>This sounds correct. All of this falls under theterm "arbitrary."

> The reals as space aren't first-countable because there are> uncountably many neighborhoods for each point (in ZFC).

No. First countability is a countable local basis aroundeach point. Since the usual topology is defined interms of the open epsilon balls strictly less than somegiven number at each point and the rational numbers areincluded among those given numbers, the usual topologyis also first countable.

> Here, each of> those (uncountably-many) delta-neighborhoods, open or closed, contains> rationals that each epsilon<delta-neighborhood does not. Then we> would be arguing about first-countability. There are uncountably many> neighborhoods of each point in topology in ZFC each contains rationals> that narrower neighborhoods do not, thus it is a contradiction that> there aren't uncountably many of those, at least a distinct one for> each.>> There are and aren't: contradiction.

I understand what you are trying to say. If itwere not the case that the order relations organizethe identity criteria in the hierarchy of logicaltypes, I might agree. But the reals are not distinctfrom their definition in terms of rationals understoodas a lower logical type.

>> Is there a neighborhood with standardly-valued radius r of a point in> R without uncountably many neighborhoods with radius < r? No.

Correct.

> Is it> so that for each ordinal gamma > beta that for diminishing values of r> through uncountably many ordinals that r_gamma contains irrationals> not in r_beta?

Does such a sequence scatter the order topology inheritedfrom the rationals of lower logical type? By this, Imean that the usual topology is defined with respect tothe inherited order and not by order-types associatedwith ordinal sequences. At r_gamma you could have avery small radius but at r_gamma+1 you could have a hugeradius.

> Directly, yes, circuitously, no. (This is where for> each r_beta there exists uncountably many values r_gamma that have via> Choice one of them, quantification over them.)>> So, are there:> a) uncountably many neighborhoods, some without distinct rationals?> b) countably many neighborhoods?> c) some other option consistent with the real space in topology and> sets of numbers in ZFC?> d) other?>

Other. Dependency on order inherited from the naturalnumbers through hierarchy of logical types.

And, importantly, relative to this there is the algebraicnotion of a norm on the rational numbers as an inner productspace. When an inner product space is transformed into aEuclidean point space, this is very similar to a metricbut not the same.

> Because, to be consistent: _all_ the properties hold _all_ the time.>Sadly, it is such a huge jigsaw puzzlethat we cannot seem to know Con(ZFC).

I know that is not what you meant...

> Countable additivity, of non-finite differentiable regions, is enough> for the integral calculus and its results that agree with geometry.

Yes. But Lebesgue measure has more to saytoward the liberties taken by Vieta as opposedto the liberties taken by Descartes.